$$f(x,y)=\begin{Bmatrix} \frac{x+y}{x^2+y^2},(x,y)\neq (0,0)) \\0 , (x,y)=(0,0)) \end{Bmatrix}$$

Let's say I have this problem. I use $y = mx$, solving this I get $\frac{1+m}{x(1+m^2)}$. Calculating the limit I get $\frac{1+m}{0}$. Could this prove that the limit doesn't exist?

Some follow-up questions:

  • When can I use th $y=mx$ method, if I can?
  • What is the rule of paths? I guess I can't just plug in $(1,2)$ or $(2,3)$, but can I plug in let's say $(\frac{1}{n}, \frac{1}{n})$? If yes, what should I look for when plugging in values? The 'n' has to go, but how?
  • If I need to prove the continuity of a function, what is the simplest way to prove it? I heard about the Squeeze Theorem, but does that work outside of sin, cos functions? At seminars we used $a^2 + b^2 >= 2ab$, but I don't find that at all clear.


$$f(x,y)=\begin{Bmatrix} \frac{x^2y3}{x^2+y^2},(x,y)\neq (0,0)) \\0 , (x,y)=(0,0)) \end{Bmatrix}$$

I have this function. Using $a^2 + b^2 >= 2ab$:

$$x^2 + y^2 >= |x^2| + |y^2| >= 2|x||y|$$

$|f(x,y)| = \frac{x^2 |y^3|}{x^2 + y^2} = \frac{x^2 |y^3|}{2|x||y|} = \frac{x y^2}{2}$ ---> this tending to 0 would prove continuity? if yes, does this method work all the time?


$$f(x,y)=\begin{Bmatrix} \frac{3x^2y}{2x^2+5y^2},(x,y)\neq (0,0)) \\0 , (x,y)=(0,0)) \end{Bmatrix}$$

$$|f(x,y)| = \frac{3x^2|y|}{2x^2+5y^2} <= \frac{3x^2|y|}{2\sqrt{10}|x||y|} = \frac{3}{2\sqrt{10}}*x$$

$$(a^2 + b^2 >= 2ab) ==> 2x^2 + 5y^2 >= (|x|\sqrt{2})^2 + (|y|\sqrt{5})^2 >= 2\sqrt{10}*x$$

$f(x,y) <= \frac{3}{2\sqrt{10}}*x$(1)

$x --> 0, y --> 0$ (2)

(1)(2) --> f(x,y) --> 0 = f(0, 0).


1 Answer 1

  1. Given a function $f$ on $\mathbb{R}^2$, it does not make sense to say "the limit" of $f$: you should indicate the limit at what point.

  2. I don't know how you get $\frac{1+m}{0}$. But restricting on $y=x$, you have $f(x,y)=\frac{1}{x}$, which tells you that the limit $\displaystyle \lim_{(x,y)\to(0,0)}f(x,y)$ does not exist.

  3. In real analysis, there is usually no one-fit-all "method". What you often have is a certain set of "strategies". You usually try the path $y=mx$ for some fixed $m$ when studying functions of the form $f(x,y)=\frac{P}{Q}$ where $P$ and $Q$ are polynomials in $x$ and $y$.

  4. Again, there is no "rule". You need to work on an ad hoc basis.

  5. "If I need to prove the continuity of a function, what is the simplest way to prove it?" There is no such thing. Sometimes, you would need to work from the first principle.

  6. In your edited question, the estimate $$ |\frac{x^2y^3}{x^2+y^2}|\le \frac{|x^2|y^3|}{2|x||y|} $$ shows that $f$ is continuous at $(0,0)$.

  • $\begingroup$ thank you, I know my terminology is a tad off, sorry for that. Yes, I meant continuous at (0, 0), and looking for limits at (0, 0). $\endgroup$
    – samivagyok
    Jan 12, 2021 at 14:28
  • $\begingroup$ @samivagyok: no problem! I was just trying to make it precise. :-) $\endgroup$
    – user9464
    Jan 12, 2021 at 14:29
  • $\begingroup$ @samivagyok Anything helpful for you here? $\endgroup$
    – user9464
    Jan 12, 2021 at 14:30
  • $\begingroup$ Yes, it was. So if I need to find continuity in a point let's say (0, 0), can I use y = mx, y = x etc.? And at the edited segment was the calculation correct? $\endgroup$
    – samivagyok
    Jan 12, 2021 at 15:06
  • $\begingroup$ @samivagyok: not quite, in general, you basically work with the definition.// for this particular problem, you work with $y=mx$ for some $m$; for your edited question, no// it should be useful to see more examples. $\endgroup$
    – user9464
    Jan 12, 2021 at 15:07

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